Question
In a $\triangle\text{ABC}$, If $2\angle\text{A}=3\angle\text{B}=6\angle\text{C}$, calculate$\angle\text{A}$,$\angle\text{B}$ and$\angle\text{C}$.
✓
Answer
In a $\triangle\text{ABC}$, $2\angle\text{A}=3\angle\text{B}=6\angle\text{C}=1$ (Suppose) $\therefore\angle\text{A}=\frac{1}{2},\angle\text{B}=\frac{1}{3}$ and $\angle\text{C}=\frac{1}{6}$$\therefore\angle\text{A}:\angle\text{B}:\angle\text{C}=\frac{1}{2}:\frac{1}{3}:\frac{1}{6}$
$=\frac{3:2:1}{6}$ But $\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$ (Sum of angles of a triangle) $\therefore\angle\text{A}=\frac{3\times180^\circ}{3+2+1}=\frac{3\times180^\circ}{6}=90^\circ$ $\angle\text{B}=\frac{2\times180^\circ}{6}=60^\circ$ $\angle\text{C}=\frac{1\times180^\circ}{6}=30^\circ$
$=\frac{3:2:1}{6}$ But $\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$ (Sum of angles of a triangle) $\therefore\angle\text{A}=\frac{3\times180^\circ}{3+2+1}=\frac{3\times180^\circ}{6}=90^\circ$ $\angle\text{B}=\frac{2\times180^\circ}{6}=60^\circ$ $\angle\text{C}=\frac{1\times180^\circ}{6}=30^\circ$
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